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This question explains why the derivative of the volume of a sphere is equal to its surface area. This is based on the logic using the differential.

However, to what extent (and when can this be generalized). For example

A. The derivative of the volume of a cube with side s does not equal its surface area.

B. The derivative of the volume of a cube with side 2r does equal its surface area.

What is going on here under the hood? Can anyone explain without using Maths beyond Calc I/II.

Also, I am not so interested in understanding why the derivative of the area is equal to the circumfence, as in understanding which shapes in general exhibit this property and why.

Note: This paper explains this question, but is quite technical. Summarizing the main results in less technical language would be useful.

Starlight
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    Could you try draw some pictures What does a sphere look like? How does the increase in volume of sphere look like ? – tryst with freedom Aug 23 '21 at 06:44
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    when you expand a cube's side from $s$ to $s + \epsilon$, the surface move a distance $\frac{\epsilon}{2}$ instead of $\epsilon$. If you want the derivative of volume equals to $A$, you need to use a parameterization such that if $t \to t + \epsilon$, the surface moves outward for a distance $\epsilon$. – achille hui Aug 23 '21 at 06:45
  • @achillehui thanks that was very useful. if you write this in an answer, I can mark it as accepted. – Starlight Aug 23 '21 at 06:59
  • One can generalize this significantly to any dimension. Suppose we have a volume which is "smoothly parameterized" by surfaces. Then, we can relate the change in volume with respect to the parameter (i.e a derivative of some kind) with an appropriate surface integral (formulating this precisely is technical, and I wrote an answer before which I may try to simplify later). In the case of a parametrization of balls using diameters, it would introduce a constant factor in the surface integral. For parametrization with radius, the constant is 1 so derivative of volume is the surface area. – peek-a-boo Aug 23 '21 at 07:23
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    Does this answer your question? Why is the derivative of a circle's area its perimeter (and similarly for spheres)? (The more general question here may not be an exact duplicate, but is addressed in the answers.) – Andrew D. Hwang Aug 23 '21 at 09:00
  • @AndrewD.Hwang This question is useful, but focuses on why the derivative has certain behaviour. I am interested in when (for which shapes) is this behaviour demonstrated. – Starlight Aug 23 '21 at 09:29
  • I ought to have linked the specific relevant answer, which gets at the shapes for which this type of relationship holds, and why. – Andrew D. Hwang Aug 23 '21 at 15:12

1 Answers1

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When built upon a single symmetric length parameter L (sphere radius, side of cube or any Platonic solid radius icosahedron &c.,) we could generally say:

$$ V=aL^3;\; A= b L^2;\;\dfrac{V^2}{A^3}= \dfrac{a^2}{b^3} = \text{constant} \tag 1 $$

By logarithmic differentiation

$$ \dfrac{2\; dV}{V}=\dfrac{3\; dA}{A}\;$$

$$\dfrac{dV}{dA}=\dfrac{3 V}{2 A} \tag2, $$

a general relation like for a cube, sphere or dodecahedron, ( that always works out in terms of their characteristic lengths $L =(a/2, r/2, R/2)$ respectively.)

EDIT1:

Can be differentiated also this way $$ V= a L^3,\dfrac{dV}{dL}=3 a L^2 $$

$$ A= b L^2, \dfrac{dA}{dL}=2 b L $$

Divide directly $$\dfrac{V}{A}=\dfrac{ A L }{b} \tag 3 $$ Divide differentials

$$\dfrac{dV}{dA}=\dfrac{3 A L }{2b} \tag 4 $$

From (3) and (4)

$$\dfrac{dV}{dA}=\dfrac{3 V}{2 A} \tag {5=2}. $$

Narasimham
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  • Which variable are you doing the logarithmic differentation with respect to? – Starlight Aug 25 '21 at 06:17
  • We can use a parameter $L$ and eliminate it, as shown in the beginning itself; it is more convenient and relevant to your question about generality to consider differentials directly without a parameter. They are shown related to $L$ in the last line. – Narasimham Aug 25 '21 at 07:04
  • If you eliminate the parameter, what are you differentiating with respect to? – Starlight Aug 26 '21 at 10:34
  • Please see the edit. Even if we differentiate at the beginning with respect to $L,$ the effect is the same end result – Narasimham Aug 26 '21 at 15:36